 So we have this question where this galvanic cell is given, these half reactions are given and their corresponding standard reduction potentials are given and the question is to calculate the E cell at a temperature of 30 degree Celsius. So you can pause the video here and give it a try before we continue. So the first thing to notice is that this temperature here is not the standard temperature that we use to calculate the E0 which is 25 degree Celsius. So to calculate the cell potential at a temperature different from the standard temperature, we use the Nernst equation which is written like this where E0 cell is the standard cell potential, R is the universal gas constant, T is the temperature and Kelvin, NF is the charge transferred and Q is the reaction quotient. So to calculate the cell potential at 30 degree Celsius, first we need to calculate the standard cell potential. So we know that the standard cell potential is given as the E0 of cathode minus the E0 of anode which we know from here because this galvanic cell is written in alphabetical order. So we know that this is the anode, this is the bridge and this is the cathode. So the standard potential of the cathode which is given here is this positive 0.8 volt minus the standard potential of the anode which is negative 1.66 volts. So if we solve this, we get the standard cell potential to be equal to 2.46 volts. So we know this value. Now before we calculate the next part, let's write down the half cell reactions like this. So on the cathode part which is on the right side, we have the reduction half reaction which is written like this and on the anode side, we have the oxidation half reaction which is written like this. So to get the cell reaction, because we have three electrons here and one electron here, we will multiply this reaction by three and then we add both of these half reactions. So we get this as the cell reaction. Now looking at the cell reaction, we know that here the number of electrons that have transferred that is the coefficient here is 3. So if we look at these values, R is the universal gas constant and we know its value. T is the temperature in Kelvin and because this is at 30 degrees, for Kelvin, we will add 273 to it. So we get the temperature to be 303 Kelvin. N is the number of electrons transferred which we know from this is 3 because we multiply 3 here and F is the Faraday constant and its value is given here. So now we know the values of R, T, N and F. So all that is left is to get the value of this Q. So this Q is the reaction quotient and we are going to get this from our cell reaction. So to get this Q, we are going to use our cell reaction and for this reaction, we can write the reaction quotient like this which is the concentration of the Al3 plus iron which is on the product side divided by the concentration of this Ag plus iron raised to the power 3 which is the coefficient here and here the concentration of the other species like this silver or this aluminum will be equal to 1 because in case of a pure solid or a pure liquid, we take their activities or concentrations to be 1 because in case of pure solids and pure liquids, the molar concentration is proportional to density and the density for them does not change in the reaction and since the concentration for both of these is 1, I have not written them down here and for these two, the values are already given here. So if we substitute these values, we get the reaction quotient to be 0.15 divided by 1.55 to the power 3. So now all that is left to calculate this value of E cell is to plug in all these values. The standard cell potential which is 2.46 volts, R, T, N, F and this Q. Sometimes the value of this ln Q is directly given in the question. So in that case, you don't need to write down the half reactions and also notice that the function here is of natural log and not log base 10 and if you plug in everything into a calculator and if you haven't made any mistakes in calculation, you will get the final answer of the cell potential to be 2.49 volts.